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2 Aug 2018: Chern number can be obtained by tomography of Bloch statesTwo-band model: |uk (t)〉 = sin(θk/2)|A, k〉 cos(θk/2) exp(iφk )|B, k〉. ... N. Fläschner, B. S. Rem, M. Tarnowski,. D. Vogel, D.-S. Lühmann, K.

1 Nov 2018: CA =. 0. B@cos sin 0sin cos 0. 0 0 1. ... X cos Y sin. Again, U () must act on X this way if it is indeed to correspond to arotation.

13 Dec 2018: A cos nθ B sin nθ n 6= 0(2.22). Returning to (2.18),. ... Hence, we obtain possible solutions to (2.17) as. Φ = RΘ =. . (C D ln r)(A Bθ) n = 0. (Crn Drn)(A cos nθ B sin nθ) n 6=

13 Dec 2018: Example: if a and b are vectors then a. b is a scalar. ... b/|a| |b| = cos θ is a scalar, so that the angle between vectors is unaffected by a changeof frame.

28 Nov 2018: A group (G, ,e) is abelian2 if for all a,b G we have a b = b a. ... f ′(f(z)) =a′(azbczd) b. ′. c′(azbczd) d′. =a′(az b) b′(cz d).

8 Jan 2018: 53. K43. Observe that. f(h) f(0)h. = |hsinh4| |h| 0. as h 0.If x 6= 0, then f ′(x) = 2xsinx4 4x2 cos x4. ... K87. Very similar to K86. To obtain the formula for un, observe that2 cos2 φ = 1 cos 2φ and cos(π/4) = 21/2.

29 Apr 2018: Ceci prouve que A A, A Y σ(L). De même, B B, X B σ(L). ... Il suffit de prouver que 1I Vect (en, n Z).On écrit I = [ch,c h] et on pose :. ϕ : t [a,b] 7 12 [cos (ω(t c)) cos(ωh)].

13 Mar 2018: divv = (ux vy) (3). curlv = (uy vx) (4). (defv) cos 2µ = (ux vy) (5). ... Z0(56). curlv = 23 (qU1 pU2). Z0(57). (defv) cos 2µ =(pU1 qU2).

13 Mar 2018: Input. v2. u2. θ2. Online Update. P1. P2. Manifold Learning. θΣby cos i. ... b) the low-dimensional manifoldrepresentation allows matching that is both time and memoryefficient.

13 Mar 2018: divv = (ux vy) (3). curlv = (uy vx) (4). (defv) cos 2µ = (ux vy) (5). ... Z0(56). curlv = 23 (qU1 pU2). Z0(57). (defv) cos 2µ =(pU1 qU2).

13 Mar 2018: Input. v2. u2. θ2. Online Update. P1. P2. Manifold Learning. θΣby cos i. ... b) the low-dimensional manifoldrepresentation allows matching that is both time and memoryefficient.

3 Apr 2018: 1/C0 sin(. C0κ1) cos κ3φ2. /C0 sin(. C0κ1) sin κ3φ1. ]J2 cos(. ... Aκ1 = φ2J1 φ1J2. Aκ2 =1C0. sin(C0κ1)φ1J1. 1C0. sin(C0κ1)φ2J2+. cos(C0κ1)J3. A direct calculation shows that this gauge potential is indeed GC0-equivariant.

17 Aug 2018: cos(0) = 1).We then define B to be the symmetric bilinear form on V = RS whose matrix in the canonicalbasis (es)sS is C. ... Y ) ((t,y) , (t′,y′))2 = t2 t′2 2tt′ cos (min{π,dY (y,y′)}).

22 Nov 2018: systemic velocity of the galaxyindicated by the Doppler shift of the nucleus, and b/a = cos i is the ratioof the minor to the major axis of the ellipse that a ... Right : The observer sees distorted, correlated, images of the distant galaxies.The

13 Mar 2018: S 1cosÿ. ÿcos ÿ2 sin cos 2d cos ÿ2 sin cos cos 2d sin. ... transformation is further reduced to a bilateral symmetry,. given by. B ÿcos 2 ÿsin 2 2d cos ÿsin 2 cos 2 2d sin.

13 Mar 2018: 0 0 cos(φ θ).  , (14). wheredl = u0 cos θ v0 sin θ. ... will then become abilateral symmetry, given by. B =. . cos 2θ sin 2θ 2dl cos θ sin 2θ cos 2θ 2dl sin θ.

16 Nov 2018: ThenAzz̄ Bz̄ B̄z C = 0 |z B/A|2 = |B|2/A2 C/A. ... For z C, z = f(z) if and onlyif cz2 dz az b = 0.

13 Mar 2018: S 1cosÿ. ÿcos ÿ2 sin cos 2d cos ÿ2 sin cos cos 2d sin. ... transformation is further reduced to a bilateral symmetry,. given by. B ÿcos 2 ÿsin 2 2d cos ÿsin 2 cos 2 2d sin.

14 Jun 2018: transit,. tcross(b = 0) =(. 2R?G M? ) rt/(1+ e cos $). ... rt < 0.31(1+ e cos $) (22). in au, and so rt < 0.62 au.

1 Nov 2018: 122 //======================================== Find axis ranges & change B-factors if required.

13 Mar 2018: 0 0 cos(φ θ).  , (14). wheredl = u0 cos θ v0 sin θ. ... will then become abilateral symmetry, given by. B =. . cos 2θ sin 2θ 2dl cos θ sin 2θ cos 2θ 2dl sin θ.

12 Jul 2018: regularisations on sigmoid stimulated DNNs. 1135.6 WSJ-SI84: Comparison of activation grid outputs of raw, KL, Cos and. ... That is, suppose a simple linear function is applied2:. φ(z) = W Tz b.

13 Mar 2018: 1B = cos(b1) sin(b1) b2sin(b1) cos(b1) b3. 0 0 1.  ,1C =.  c1 cos(c2) c1 sin(c2) c3c1 sin(c2) c1 cos(c2) c40 0 ... Figure 7. Rectification of uncalibrated images. (a) and (b): Thefigures show two images.

13 Mar 2018: 1B = cos(b1) sin(b1) b2sin(b1) cos(b1) b3. 0 0 1.  ,1C =.  c1 cos(c2) c1 sin(c2) c3c1 sin(c2) c1 cos(c2) c40 0 ... Figure 7. Rectification of uncalibrated images. (a) and (b): Thefigures show two images.